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1、一些關于二進制轉換十進制的代碼,在網(wǎng)上看到的,總結了一下,第一個是我自己寫的,其他的是別人的.其中有一些很經(jīng)典,有一些很復雜,但是效率很高,有一些我看不懂的.1.#includeusingnamespacestd;classBinarySystem{public:BinarySystem(intnumber);intbCout();private:intrNum;};BinarySystem::BinarySystem(intnumber){rNum=0;intresult(number),residue(
2、0);while(result!=1){result=result/2;residue=result%2;if(residue==1)++rNum;}}intBinarySystem::bCout(){rNum+=1;returnrNum;}intmain(){cout<<"Pleaseinsertanumber"<>a;BinarySystembinary(a);b=binary.bCout();cout<<"Thenumberis"<3、intbit_count(unsignedintn){intcount;for(count=0;n;n&=n-1)count++;returncount;}voidmain(){unsignedintn;intcnt;cout<<"Inputyournum:"<>n;cnt=bit_count(n);cout<<"countnumofbits:"<4、+1))#defineROUND(n,c)(((n)&MASK(c))+((n)>>POW(c)&MASK(c)))intcount(intn){n=ROUND(n,0);n=ROUND(n,1);n=ROUND(n,2);n=ROUND(n,3);n=ROUND(n,4);returnn;}4.#include//Hacker'sDelight,5-1//Figure5-2Counting1-bitsinaword.intpop(unsignedx){x=x-((x>>1)&0x55555555);x=(x
5、&0x33333333)+((x>>2)&0x33333333);x=(x+(x>>4))&0x0F0F0F0F;x=x+(x>>8);x=x+(x>>16);returnx&0x0000003F;}intmain(){unsignedlongx;while(1){printf("inputaninteger:");scanf("%d",&x);if(x==0)break;printf("0x%08X:Numberof1-bits:%d",x,pop(x));}return0;}5.根據(jù)十一樓的想法編寫的:一個奇數(shù)/2后的
6、二進制數(shù)1個數(shù)+1就得到原來的二進制數(shù)intbit_count(inta){intb=0;while(a){if(a%2)b++;a=a/2;}returnb;}